Bounded sequence and every convergent subsequence converges to L [duplicate]

Solution 1:

I revise your proof.

Let {$x_n$} be bounded, and every subsequence converges to L. Assume that $lim_{n\to\infty}(x_n)\ne L$. Then there exists an epsilon such that infinitely many $n \in N \implies |x_n - L|\ge \epsilon $ Now, there exists a subsequence $\{ x_{\Large{n_k}} \}$ such that $|x_{\Large{n_k}} - L|\ge \epsilon \quad \color{red}{(♫)}$

1. How to presage proof by contradiction? Why not a direct proof?

2. Where does $\color{red}{(♫)}$ issue from?

By Bolzano Weiertrass Theorem $\{ x_{\Large{n_k}} \}$ has a convergent subsequence $\{ x_{n_{k_l}} \}$ that doesn't converge to L. This is a contradiction.

Why? $\{ x_{\Large{n_{k_l}}} \}$ is a sub sequence of the sub sequence $x_{\Large{n_k}} $, which was posited to converge to L.
By the agency of p 57 q2.5.1, every convergent sub sequence of $x_n$ converges to the same limit as the original sequence. So $\{ x_{\Large{n_{k_l}}} \} \to L$.